Question

A block of mass $m=2$ kg is suspended with a light spring of spring constant $k=1000$ N/m. If the lift is moving vertically upward with acceleration of 2 m/s$^2$ then equilibrium elongation in the spring is $(g=10$ m/s$^2$)

• A. 3.2 cm
• B. 2.4 cm
• C. 1.2 cm
• D. 2.0 cm

Answer 1

Answer: (B)

2.4 cm

Answer 2

The block experiences an upward spring force and a downward gravitational force. Since the lift is accelerating upwards, the net force on the block must be upward and equal to $ma$. This leads to the equation $kx - mg = ma$. Solving for $x$ gives $x = \frac{m(g+a)}{k}$. Plugging in the given values, $x = \frac{2(10+2)}{1000} = \frac{24}{1000} = 0.024$ m, which is 2.4 cm.